Noether Normalization

This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.

Noether Normalization

The goal of this post is to prove the following theorem and examine its consequences.

Theorem 1 (Noether normalization lemma) Let \(A\) be a finitely generated \(d\)-dimensional \(\mathbb{K}\)-algebra. Suppose natural numbers satisfying the inequality

\[d_1>d_2>\cdots>d_m>0\]

and a descending chain of ideals of \(A\)

\[\mathfrak{a}_1\subset \mathfrak{a}_2\subset\cdots\subset \mathfrak{a}_m\]

satisfying \(\dim \mathfrak{a}_i=d_i\) are given. Then there exists a suitable subring \(B\cong \mathbb{K}[\x_1,\ldots, \x_d]\) of \(A\) such that \(A\) is finitely generated as a \(B\)-module and we can arrange that the following equality

\[\mathfrak{a}_i\cap B=(\x_{d_i+1},\ldots, \x_d)\qquad\text{for $i=1,\ldots, m$}\]

holds.

This can be proved using the following lemma, and we omit its proof.

Lemma 2 Let \(\mathbb{K}\) be a field and let \(f\in B=\mathbb{K}[\x_1,\ldots, \x_r]\) be a non-constant polynomial. Then there exist suitable elements \(\x_1',\ldots, \x_{r-1}'\in B\) such that, letting \(B'\) be the \(\mathbb{K}\)-subalgebra of \(B\) generated by \(\x_1',\ldots, \x_{r-1}', f\), we can arrange that \(B\) is a finitely generated \(B'\)-module. Moreover, these elements can be chosen as follows.

1. For a sufficiently large integer \(e\), we can choose \(\x_i'=\x_i-\x_r^{e}\).
2. If \(\mathbb{K}\) is an infinite field, then for suitable \(a_i\in \mathbb{K}\) we can choose \(\x_i'=\x_i-a_i\x_r\).

Then the proof of Theorem 1 is as follows.

Proof of Theorem 1

Since \(A\) is a finitely generated \(\mathbb{K}\)-algebra, we can write \(A=\mathbb{K}[\y_1,\ldots, \y_r]/\mathfrak{a}\). Then, given a chain of ideals satisfying the conditions, consider the chain consisting of their preimages in \(\mathbb{K}[\y_1,\ldots, \y_r]\)

\[\tilde{\mathfrak{a}}_1\subset \tilde{\mathfrak{a}}_2\subset\cdots\subset \tilde{\mathfrak{a}}_m\]

and insert \(\mathfrak{a}_0=\mathfrak{a}\) to view it as a descending chain of ideals in \(\mathbb{K}[\y_1,\ldots, \y_r]\)

\[\mathfrak{a}\subset \tilde{\mathfrak{a}}_1\subset \tilde{\mathfrak{a}}_2\subset\cdots\subset \tilde{\mathfrak{a}}_m\]

so it suffices to prove the claim for the polynomial ring \(A=\mathbb{K}[\y_1,\ldots, \y_r]\). In this case, we must have \(r=d\) by §System of Parameters, ⁋Corollary 10.

To construct the elements \(\x_i\) of the theorem, we first set \(\x_i'=\y_i\), and then modify them step by step to find \(\x_d\) satisfying the given conditions. For this purpose, assume that elements \(\x_1',\ldots, \x_e', \x_{e+1},\ldots, \x_d\) satisfying the following two conditions are given:

1. \(A\) is a finitely generated \(B_e=\mathbb{K}[\x_1',\ldots, \x_e',\x_{e+1},\ldots, \x_d]\)-module.
2. For each \(i\), the inclusion \(\mathfrak{a}_i\cap B_e\supset(\x_m,\ldots, \x_d)\) holds, where \(m=\max(d_i+1, e+1)\).

We show that from these we can find new elements \(\x_1',\ldots, \x_{e-1}'\) and \(\x_e\) so that the above conditions are preserved. Then, repeating this process, it is clear that the final \(B=B_{d_m}\) satisfies the desired conditions once we show that the inclusion in the second condition is actually an equality, which is obvious by considering the dimensions of the two ideals of \(B\) on both sides.

Now, to complete this induction, suppose that for some \(e\) with \(d\geq e>d_m\), elements \(\x_1',\ldots, \x_e', \x_{e+1},\ldots, \x_d\) satisfying the two conditions above are given, and assume that \(i\) is the smallest index satisfying \(e>d_i\). Then

\[\mathfrak{a}_i\cap \mathbb{K}[\x_1',\ldots, \x_e']\neq 0\]

Indeed, if this intersection were \(0\), then by the second condition we would have

\[\mathfrak{a}_i\cap B_e\supseteq (\x_{e+1},\ldots, \x_d),\]

because the ideal on the left-hand side has dimension \(d_i\), while the ideal on the right-hand side has dimension \(e\), which is a contradiction. Now choose \(\x_e\) to be any nonzero polynomial belonging to the above intersection, and then use Lemma 2 to replace the elements \(\x_1',\ldots, \x_{e-1}'\) with new ones as well.

Consequences

Theorem 1 yields the following result.

Theorem 3 Let \(A\) be an integral domain that is a finitely generated \(\mathbb{K}\)-algebra. Then \(\dim A=\trdeg_\mathbb{K}\Frac(A)\).